Theorem: If $W$ is a set of one or more vectors in a vector space $V$, then $W$ is a subspace of $V$ if and only if the following conditions are satisfied.
- If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $W$, then $\mathbf{u+v}$ is in $W$.
- If $k$ is a scalar and $\mathbf{u}$ is a vector in $W$, then $k \mathbf{u}$ is in $W$.
Use the theorem to determine which of the following are subspaces of $P_3$.
- All polynomials $a_0 +a_1 x + a_x x^2 +a_3x^3$ for which $a_0=0$.
- All polynomials $a_0 +a_1 x + a_x x^2 +a_3x^3$ for which $a_0+a_1+a_2+a_3=0$.
- All polynomials of the form $a_0 +a_1 x + a_x x^2 +a_3x^3$ in which $a_0,a_1,a_2,$ and $a_3$ are rational numbers.
- All polynomials of the form $a_0+a_1x$, where $a_0$ and $a_1$ are real numbers.
I'm having trouble understanding how this theorem relates to these generalized examples here..
Such as (a), I'm assuming $a_0, a_1$, etc are vectors like $u$ and $v$, and $P_3$ is $V$, but I'm not sure what $W$ is precisely, is it all polynomials that fit that description? I'm kind of forgetting how to show if the form is $P_3$ mathematically, but I'm assuming it is because it's rank $3$. But what does $a_0 = 0$ have to do with determining it?
How am I supposed to test if "all polynomials" are in $W$? Is anyone else good at these generalized vector space problems? I'm having trouble getting a good grasp of real vector spaces and subspaces for answering relative problems…
Best Answer
Let's try to breakdown understanding in several steps for clarification.
With this in mind, let's determine if $$W_1 = \{ p \in P_3 \, : \, p(x) = a_0+a_1x + a_2x^2 +a_3x^3, \quad a_0 = 0 \}$$ is a subspace of $P_3$. Since $a_0=0$ then really all polynomials in $W_1$ are of the form $$p_1(x) = a_1x + a_2x^2 +a_3x^3.$$ Now we need to check the conditions. If we sum another polynomial $p_2 (x) = b_1 x +b_2x^2 +b_3x^3$ to $p_1(x)$, is the sum still a polynomial with null independent coefficient? If we multiply either of those by a number, will the independent coefficient still be zero? We have $$p_1(x) + p_2(x) = (a_1+b_1)x + (a_2+b_2)x^2 +(a_3+b_3)x^3 = c_1x +c_2 x^2 +c_3x^3,$$ therefore the sum is still a polynomial with zero independent coefficient. Multiplying by a number gives $$k p_1(x) = (ka_1) x + (ka_2)x^2 +(ka_3)x^3 = d_1 x + d_2x^2 +d_3x^3,$$ and it's still of the form given. Since the conditions are satisfied, we've proved $W_1$ is a subspace of $P_3$. Check the others.